6.4 Answer

6.4

1.

(a) $f_{x}(x,y) = 3x^2 y^4, f_{y}(x,y) = 4x^3 y^3$より

$$df(x,y) = 3x^2 y^4 dx + 4x^3 y^3 dy$$

$$\nabla f(x,y) = (3x^2 y^4, 4x^3 y^3)$$

The tangent plane goes through the point corresponding to $(1,1)$ is

$$z = f(1,1) + \nabla f(1,1) \cdot (x - 1,y-1)$$

$$= 1 + 3(x-1) + 4(y-1) = 3x + 4y -6$$

The normal line goes through the point corresponds to $(1,1)$ is

$$(x,y,z) - (1,1,1) = t(3,4,-1)$$

or，

$$t = \frac{x-1}{3} = \frac{y-1}{4} = \frac{z -1}{-1}$$

(b)

$f_{x}(x,y) = 3x^2 y + 2xy^4, f_{y}(x,y) = x^3 + 4x^2 y^3$ implies

$$df(x,y) = (3x^2y + 2xy^4) dx + (x^3 + 4x^2 y^3) dy$$

$$\nabla f(x,y) = (3x^2y + 2xy^4, x^3 + 4x^2 y^3)$$

The tangent plane goes through the point corresponds to $(1,1)$ is

$$z = f(1,1) + \nabla f(1,1) \cdot (x - 1,y-1)$$

$$= 2 + (5,5) \cdot (x-1,y-1) = 2 + 5x - 5 + 5y - 5$$

$$= 5x + 5y - 8$$

The normal line goes through the point corresponds to $(1,1)$ is

$$(x,y,z) - (1,1,2) = t(5,5,-1)$$

or

$$t = \frac{x-1}{5} = \frac{y-1}{5} = \frac{z -2}{-1}$$

(c)

$z_{x} = 2xye^{2x} + 2x^2 y e^{2x}, z_{y} = x^2 e^{2x}$より

$$dz = (2xye^{2x} + 2x^2 y e^{2x}) dx + x^2 e^{2x}dy$$

$$\nabla f(x,y) = (2xye^{2x} + 2x^2 y e^{2x}, x^2 e^{2x})$$

THe tangent plane goes through the point corresponds to 点$(1,1)$ is

$$z = e^2 + (2e^2 + 2e^2) \cdot (x - 1,y-1)$$

$$= e^2 + 4e^2(x-1) + e^2(y-1)$$

The normal line goes through the point corresponds to $(1,1)$ is

$$(x,y,z) - (1,1,e^2) = t(4e^2,e^2,-1)$$

or

$$t = \frac{x-1}{4e^2} = \frac{y-1}{e^2} = \frac{z - e^2}{-1}$$

2.

(a) Consider $f(x,y) = \sqrt{x}\sqrt[4]{y}$. Then $f(121,16) = \sqrt{121}\sqrt[4]{16} = 22$．Then the seeking vlaue is $f(125,17) = f(121+4,16+1)$．Now let $\Delta x = 4, \Delta y = 1$. Then

$$f(121+4,16+1) = f(121,16) + \Delta f(121,16)$$

$$\approx f(121,16) + df(121,16)$$

$$df = f_{x}dx + f_{y}dy$$

$$= \frac{1}{2}x^{-\frac{1}{2}}y^{\frac{1}{4}}dx + \frac{1}{4}x^{\frac{1}{2}}y^{-\frac{3}{4}}dy$$

$\Delta x = dx, \ \Delta y = dy$ Then

$$df(121,16) = \frac{4}{11} + \frac{11}{32} \approx 0.7$$

Therefore， $f(125,17) \approx 22.7$

(b) Consider $f(x,y) = \sin{x}\cos{y}$. Then $f(\pi,\frac{\pi}{3}) = \sin{\pi}\cos{\frac{\pi}{3}} = 0$．The seeking value is $f(\frac{6\pi}{7},\frac{\pi}{3}) = f(\pi - \frac{\pi}{7},\frac{\pi}{3})$．Now let $\Delta x = -\frac{\pi}{7}, \Delta y = 0$. Then

$$f(\frac{6\pi}{7},\frac{\pi}{3}) = f(\pi,\frac{\pi}{3}) + \Delta f(\pi,\frac{\pi}{3})$$

$$\approx f(\pi,\frac{\pi}{3}) + df(\pi,\frac{\pi}{3})$$

$$df = f_{x}dx + f_{y}dy$$

$$= \cos{x}\cos{y}dx - \sin{x}\sin{y}dy$$

$\Delta x = dx, \ \Delta y = dy$ Thus,

$$df(\pi,\frac{\pi}{3}) = \cos{\pi}\cos{\frac{\pi}{3}}dx - \sin{\pi}\sin{\frac{\pi}{3}}dy =\frac{\pi}{14} \approx 0.22$$

Therefore， $f(\frac{6\pi}{7},\frac{\pi}{3}) \approx 0.22$